Skip to content

Phasor Addition

Copy this iframe to your website:

1
<iframe src="https://dmccreary.github.io/signal-processing/sims/phasor-addition/main.html" width="100%" height="612px" scrolling="no"></iframe>

Run MicroSim in Fullscreen

Description

This MicroSim shows that adding sinusoids of the same frequency is the same as adding vectors. Each sinusoid \(A_k\cos(\omega_1 t + \phi_k)\) is represented by its phasor \(A_k e^{j\phi_k}\), and

\[\sum_k A_k\cos(\omega_1 t + \phi_k) = \mathrm{Re}\Big\{\Big(\sum_k A_k e^{j\phi_k}\Big)e^{j\omega_1 t}\Big\} = |R|\cos(\omega_1 t + \angle R).\]

Engineers write \(j\) for \(\sqrt{-1}\), the \(i\) of Euler's formula. The left panel uses the colours of the Euler's Formula Explorer: a blue real axis, a red imaginary axis and a green vector, here the resultant \(R = \sum_k A_k e^{j\phi_k}\). Up to three phasors (orange, purple, brown) are drawn head to tail and turn together at \(\omega_1 = 2\pi f_1\). The blue dot is the resultant's projection on the real axis, which is the value of the sum at that instant. The right panel plots each waveform as a thin line, the sum as a thick green line, the envelope \(\pm|R|\) as dashed lines, and a gold cursor at the current time over an 8 s window.

Each phasor has a checkbox, an amplitude slider (0 to 2) and a phase slider (−180° to 180°). Speed sets \(f_1\) (0.10 to 2.00 Hz). The MicroSim opens paused at \(t = 0\), so each arrow sits at its phase angle \(\phi_k\); Play starts the rotation, Pause stops it, and Reset returns to \(t = 0\) and pauses. The read-out gives \(R\) in polar and rectangular form. Phasor 2 at f₂ makes phasor 2 turn at a second frequency \(f_2\). The resultant's length then changes with time, the dashed envelope becomes \(\pm|R(t)|\), and the read-out gives the beat frequency \(|f_2 - f_1|\).

This simulation is used in Chapter 1: Mathematical Foundations.

Lesson Plan

Learning Objectives

After using this MicroSim, students will be able to:

  1. Calculate the resultant phasor \(R = \sum_k A_k e^{j\phi_k}\) for up to three sinusoids of one frequency and write their sum as \(|R|\cos(\omega t + \angle R)\)
  2. Predict from the phase difference whether two sinusoids reinforce, partly cancel or cancel completely, using \(|R| = 2A|\cos(\Delta\phi/2)|\) for equal amplitudes
  3. Demonstrate with the speed slider that the resultant amplitude and phase do not depend on the rotation rate \(\omega\) when every phasor shares that frequency
  4. Relate the projection of the rotating resultant onto the real axis to the value of the summed waveform at the cursor
  5. Analyze the sum of two sinusoids at different frequencies as a resultant whose length changes with time, and calculate the beat frequency \(|f_2 - f_1|\) and the envelope range

Target Audience

Undergraduate students in a first signals and systems or circuits course. Plan about 25 minutes for the activities.

Prerequisites

  • Complex numbers in rectangular and polar form (see the Complex Plane MicroSim)
  • Euler's formula \(e^{j\theta} = \cos\theta + j\sin\theta\)
  • Amplitude, frequency and phase of a sinusoid

Activities

Activity 1: Two phasors at right angles (6 min). Start from the default settings: phasor 1 is \(1.00\angle 0°\), phasor 2 is \(1.00\angle 90°\), phasor 3 is off and \(f_1 = 0.50\) Hz. Press Reset: arrow 1 lies on the real axis and arrow 2 points straight up from its tip. The read-out shows \(R = 1.41\angle 45.0°\) \(= 1.00 + j1.00\), and the thick green sum peaks at 1.41, not 2: the sum is \(1.41\cos(\omega_1 t + 45.0°)\). At \(t = 0\) the blue dot sits at 1.00, the same value as the green dot on the cursor. Press Play and check that the two stay equal. Then move the speed from 0.50 Hz to 1.50 Hz: the plot shows 12 cycles in the 8 s window instead of 4 and \(\omega_1\) rises from 3.14 to 9.42 rad/s, but \(R\) stays \(1.41\angle 45.0°\).

Activity 2: The phase difference decides (5 min). Keep \(A_1 = A_2 = 1.00\) and \(\phi_1 = 0°\), and set \(\phi_2\) to each value below. Compare each read-out with \(|R| = 2\cos(\phi_2/2)\):

  • \(\phi_2 = 0°\): \(R = 2.00\angle 0.0°\), the amplitudes add
  • \(\phi_2 = 60°\): \(R = 1.73\angle 30.0°\)
  • \(\phi_2 = 120°\): \(R = 1.00\angle 60.0°\), the sum is no larger than either input
  • \(\phi_2 = 180°\): the read-out says \(R = 0\) and the green sum is flat

Activity 3: A balanced three-phase set (5 min). Check Phasor 3 and set all three amplitudes to 1.00 with \(\phi_1 = 0°\), \(\phi_2 = 120°\) and \(\phi_3 = -120°\). The three arrows close an equilateral triangle, \(R = 0\), and the sum is flat although each waveform has amplitude 1. This is why the currents of a balanced three-phase load cancel in the neutral wire. Now set \(A_3 = 0.50\): the triangle no longer closes and \(R = 0.50\angle 60.0°\) \(= 0.25 + j0.43\).

Activity 4: Beats (7 min). Set \(A_1 = A_2 = 1.00\), \(\phi_1 = \phi_2 = 0°\), phasor 3 off and \(f_1 = 1.00\) Hz. Check Phasor 2 at f₂ and set \(f_2 = 1.25\) Hz. Phasor 2 now gains a quarter turn per second on phasor 1, so the length of \(R\) swings between 0.00 and 2.00. The read-out gives a beat frequency of 0.25 Hz and a beat period of 4.00 s. In the first window the dashed envelope \(2|\cos(\pi \cdot 0.25\,t)|\) touches zero at \(t = 2\) s and \(t = 6\) s, and the sum reaches 2 at \(t = 0\), 4 and 8 s. Then set \(A_2 = 0.50\): \(|R(t)|\) now swings between 0.50 and 1.50 and the sum never dies out.

Assessment

  1. Write \(1.5\cos(\omega t) + 2\cos(\omega t + 90°)\) as a single sinusoid. Answer: \(R = 1.5 + j2.0 = 2.50\angle 53.1°\), so the sum is \(2.50\cos(\omega t + 53.1°)\). Check it with \(A_1 = 1.50\), \(A_2 = 2.00\), \(\phi_2 = 90°\).
  2. What is \(2\cos(\omega t + 30°) + 2\cos(\omega t - 30°)\)? Answer: \(|R| = 2 \cdot 2\cos 30° = 3.46\) and the imaginary parts cancel, so the sum is \(3.46\cos(\omega t)\).
  3. Phasors \(1\angle 0°\) and \(1\angle 90°\) are fixed. Which third phasor makes the sum zero? Answer: it must cancel \(R = 1.41\angle 45°\), so \(A_3 = \sqrt{2} \approx 1.41\) at \(\phi_3 = -135°\). The slider moves in 0.05 steps; \(A_3 = 1.40\) leaves \(|R| = 0.01\).
  4. Two tones of amplitude 1 and zero phase have frequencies 0.80 Hz and 1.00 Hz. What are the beat frequency and the beat period, and when does the sum first vanish? Answer: 0.20 Hz and 5.00 s. The envelope \(2|\cos(\pi \cdot 0.20\,t)|\) is first zero at \(t = 2.5\) s, and again at 7.5 s.
  5. Why can't the two tones in question 4 be replaced by one phasor? Answer: the angle between them grows by \(2\pi(f_2 - f_1)t\), so their vector sum has no fixed length. Its length, the envelope, varies between \(|A_1 - A_2|\) and \(A_1 + A_2\).

Common Misconceptions

"The amplitudes of two sinusoids add." Only phasors that point the same way add their lengths. At the default settings \(1.00\angle 0°\) plus \(1.00\angle 90°\) gives a sum with amplitude 1.41. Only \(\phi_2 = 0°\) gives 2.00.

"The phase of the sum is the average of the phases." That is true only for equal amplitudes, as at the default settings (\(45°\) halfway between \(0°\) and \(90°\)). Set \(A_1 = 2.00\): the read-out shows \(R = 2.24\angle 26.6°\), because the longer arrow pulls the resultant toward itself.

References

  1. Phasor - Wikipedia - phasor notation, phasor addition and the link to Euler's formula
  2. Euler's formula - Wikipedia - why \(A\cos(\omega t + \phi) = \mathrm{Re}\{Ae^{j\phi}e^{j\omega t}\}\)
  3. Beat (acoustics) - Wikipedia - interference of two tones and the beat frequency \(|f_1 - f_2|\)
  4. Three-phase electric power - Wikipedia - balanced phasors 120° apart that sum to zero
  5. Complex number - Wikipedia - rectangular and polar forms used in the read-out
  6. The Scientist and Engineer's Guide to Digital Signal Processing - Steven W. Smith - free online textbook with a chapter on complex numbers